# Rolle's Theorem Question Help

Given an interval $[a,b]$ that satisfies hypothesis of Rolle's theorem for function $$f(x) = x^4 + x^3 - x^2 + x -2$$

If $a = -2$, how do I find $b$ ?

This is what Ive done so far, Not sure if it is right...

Rolles theorem says if $f(a)=f(b)$ at some point in the interval $[a,b]$ the derivative of the function is zero.

so,$f(a)=32-8-8-4=12$

Lets find a $b$ where $f(b)=12$

so, $f(b)=b^4+b^3-b^2+b-2=12$

This is hard to do, but wolfram alpha says $b=1.77$

So somewhere between $-2$ and $1.77$ the derivative is zero.

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Yes. Not sure what your question is. You can also just double check your own work by taking the derivative $4x^3+3x^2-2x+1=0$, which again by wolfram alpha, we see the derivative is zero at $~-1.28$ which is between $-2$ and $1.77$ – Daniel Montealegre Apr 2 '12 at 7:41
I think you should check your calculations, $f(a) = f(-2) \neq 12$ ... – in_wolframAlpha_we_trust Apr 2 '12 at 7:45
Once you have the correct $f(a)$, you might want to check $b=1$. – Did Apr 2 '12 at 8:42

Since $f$ can be rewritten as $f(x) = (x-1)(x+2)(x^2+1)$, you can see that $f(-2) = f(1)$. There is a local extremum somewhere in $(-2,1)$.